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Brums [2.3K]
3 years ago
10

How to solve -3(1+6r)=14-r

Mathematics
1 answer:
ahrayia [7]3 years ago
6 0
The answer is r= -1
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Answer:

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Step-by-step explanation:

Given 3 \sqrt[4]{(x-2)^3}-4 = 20

now adding '4' on both sides, we get

3 \sqrt[4]{(x-2)^3}-4+4= 20+4

3 \sqrt[4]{(x-2)^3} = 24

i)  3 \sqrt[4]{(x-2)^3}  = <u>24</u>

cancelling '3' on both sides, we get

\sqrt[4]{(x-2)^3} = 8

ii)  \sqrt[4]{(x-2)^3}    =  <u> 8</u>

<u>iii) </u>(x-2)^{\frac{3}{4} } = 8<u></u>

<u></u>(x-2)^{\frac{3}{4} } = 2^{3}<u></u>

iv) (x-2)^\frac{3}{4} )^{\frac{1}{3} } = (2^{3})^\frac{1}{3}

now simplify on both sides,we get

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simplify we get solution is

<u>v) (x-2) </u>= 16

<u>vi) final answer is x = 16 +2 =18</u>

<u></u>

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Answer:

250 - 20x=y

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Steve's saving would be equal to his initial value, minus his expense, in essence;

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